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Banking of roads

  1. Sep 8, 2014 #1

    I was browsing a derivation for the relationship between the curvature of horizontal road curves and the angle of elevation/banking. The derivation is shown on page 4 here: http://www.cdeep.iitb.ac.in/nptel/Civil%20Engineering/Transportation%20Engg%201/14-Ltexhtml/nptel_ceTEI_L14.pdf [Broken]

    There is one thing I cannot understand. Everything revolves around centrifugal force. It is here treated as a force having the same magnitude of a centripetal force but in the opposite direction.

    Basically the derivation is done by first finding the component of the weight in the direction of the slope. The centrifugal force is then resolved and then force balance is used to equate the forces. (all in the direction parallel to the slope)

    So my questions are:

    1. If we resolve the component of weights and friction in the horizontal direction (instead of resolving the centrifugal force), we do not reach the same equation.

    2. Where does centrifugal force come into action? This force does not exist from my knowledge of circular motion. Only centripetal force exists. This returns us to 1. , why not resolve the weight and frictional forces in the horizontal direction and equate the forces.

    Thank you.
    Last edited by a moderator: May 6, 2017
  2. jcsd
  3. Sep 8, 2014 #2


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    You're a budding engineer. Show us your calculations. Remember, a car in a turn wants to keep travelling in a straight line, tangent to the curve.

    Because you are interested in the frictional force acting parallel to the road surface. If the curve is not banked at the correct angle for the maximum anticipated traffic speed, you could see cars flying off the banked road, causing injuries or deaths. That's why the bank angle is steeper at closed-circuit racing tracks, where speeds can reach 200 mph, than on regular roadways, where the posted speed limit is much lower.

    Last edited by a moderator: May 6, 2017
  4. Sep 8, 2014 #3
    What I did was:

    Let W = weight of vehicle
    N = Normal reaction at surface acting on vehicle
    F = Frictional force developed and y = coefficient of friction.
    x = Angle of inclination

    Resolving weight W perpendicular to the surface we have N = W cos x

    The frictional force F = Ny
    = y W cosx

    Horizontal component of F = F cos x
    = y W cos x cos x (Acting towards centre of circle)

    Horizontal component of N = N sin x
    = W cos x sin x (Also acting towards centre of circle)

    But centripetal force = centrifugal force in magnitude

    From circular motion theory, [tex] Centripetal force = \frac{MV^{2} }{R} = \frac{WV^{2} }{gR}[/tex]
    where R = Radius of curvature

    Thus from Newton's 2nd Law of motion

    [tex] \frac {WV^2}{gR} = y W cos x cos x + W cos x sin x [/tex]

    But the equation given in references actually point to:

    [tex] \frac {WV^2}{gR} cos x = y W cos x + W sin x [/tex]

    which they further simplify.
    Last edited: Sep 8, 2014
  5. Sep 8, 2014 #4


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    That is correct, but ...
    ... that is wrong. There is also a component of the centripetal force (times sin x) that is normal to the banked road.
  6. Sep 9, 2014 #5
    Isn't the centripetal force a resultant force, so it cannot be resolved?
  7. Sep 9, 2014 #6


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    I'm unfamiliar with forces that cannot be resolved.

    Obviously, if the vector forces A and B can be combined into a resultant R, then R can be decomposed into components of arbitrary direction. That's what the dot, or inner, product is for.
  8. Sep 9, 2014 #7
    Resolving the centripetal along the surface and equating with the frictional force and the weight (In the same direction) gives the required equation...
  9. Sep 9, 2014 #8


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    In rotating frames of reference. If you analyze the problem from the rest frame of the vehicle you have an inertial centrifugal force.
  10. Sep 9, 2014 #9
    Yes changing frames of reference helped in finding the solution. Thanks everyone.
  11. Sep 11, 2014 #10
    nuetral banking

    this any use ?

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